What is the result of dividing the polynomial $x^3 - 8$ by $x - 2?

Steps to Solve

1. Set up the polynomial long division

   Divide the polynomial $x^3 - 8$ by $x - 2$. We can write $x^3 - 8$ as $x^3 + 0x^2 + 0x - 8$ to keep track of place values. $$ \begin{array}{c|cc cc} \multicolumn{2}{r}{ } & & & \ \cline{2-5} x-2 & x^3 & +0x^2 & +0x & -8 \ \multicolumn{2}{r}{} & & & \ \cline{2-5} \end{array} $$

2. Divide the leading term of the dividend by the leading term of the divisor

   $x^3$ divided by $x$ is $x^2$. Write $x^2$ above the $x^2$ column. $$ \begin{array}{c|cc cc} \multicolumn{2}{r}{x^2} & & & \ \cline{2-5} x-2 & x^3 & +0x^2 & +0x & -8 \ \multicolumn{2}{r}{} & & & \ \cline{2-5} \end{array} $$

3. Multiply the quotient term by the divisor

   $x^2$ times $(x - 2)$ is $x^3 - 2x^2$. Write this below the dividend. $$ \begin{array}{c|cc cc} \multicolumn{2}{r}{x^2} & & & \ \cline{2-5} x-2 & x^3 & +0x^2 & +0x & -8 \ \multicolumn{2}{r}{x^3} & -2x^2 & & \ \cline{2-5} \end{array} $$

4. Subtract and bring down the next term

   Subtract $x^3 - 2x^2$ from $x^3 + 0x^2$ to get $2x^2$. Bring down the $+0x$. $$ \begin{array}{c|cc cc} \multicolumn{2}{r}{x^2} & & & \ \cline{2-5} x-2 & x^3 & +0x^2 & +0x & -8 \ \multicolumn{2}{r}{-x^3} & +2x^2 & & \ \cline{2-3} \multicolumn{2}{r}{0} & 2x^2 & +0x & \ \end{array} $$

5. Repeat the process

   $2x^2$ divided by $x$ is $2x$. Write $+2x$ next to $x^2$ above. $$ \begin{array}{c|cc cc} \multicolumn{2}{r}{x^2} & +2x & & \ \cline{2-5} x-2 & x^3 & +0x^2 & +0x & -8 \ \multicolumn{2}{r}{-x^3} & +2x^2 & & \ \cline{2-3} \multicolumn{2}{r}{0} & 2x^2 & +0x & \ \end{array} $$ $2x$ times $(x - 2)$ is $2x^2 - 4x$. Write this below $2x^2 + 0x$. $$ \begin{array}{c|cc cc} \multicolumn{2}{r}{x^2} & +2x & & \ \cline{2-5} x-2 & x^3 & +0x^2 & +0x & -8 \ \multicolumn{2}{r}{-x^3} & +2x^2 & & \ \cline{2-3} \multicolumn{2}{r}{0} & 2x^2 & +0x & \ \multicolumn{2}{r}{} & 2x^2 & -4x & \ \cline{3-4} \end{array} $$

   Subtract $2x^2 - 4x$ from $2x^2 + 0x$ to get $4x$. Bring down the $-8$. $$ \begin{array}{c|cc cc} \multicolumn{2}{r}{x^2} & +2x & & \ \cline{2-5} x-2 & x^3 & +0x^2 & +0x & -8 \ \multicolumn{2}{r}{-x^3} & +2x^2 & & \ \cline{2-3} \multicolumn{2}{r}{0} & 2x^2 & +0x & \ \multicolumn{2}{r}{} & -2x^2 & +4x & \ \cline{3-4} \multicolumn{2}{r}{} & 0 & 4x & -8 \ \end{array} $$

6. Repeat the process one last time

   $4x$ divided by $x$ is $4$. Write $+4$ next to $2x$ above. $$ \begin{array}{c|cc cc} \multicolumn{2}{r}{x^2} & +2x & +4 & \ \cline{2-5} x-2 & x^3 & +0x^2 & +0x & -8 \ \multicolumn{2}{r}{-x^3} & +2x^2 & & \ \cline{2-3} \multicolumn{2}{r}{0} & 2x^2 & +0x & \ \multicolumn{2}{r}{} & -2x^2 & +4x & \ \cline{3-4} \multicolumn{2}{r}{} & 0 & 4x & -8 \ \end{array} $$ $4$ times $(x - 2)$ is $4x - 8$. Write this below $4x - 8$. $$ \begin{array}{c|cc cc} \multicolumn{2}{r}{x^2} & +2x & +4 & \ \cline{2-5} x-2 & x^3 & +0x^2 & +0x & -8 \ \multicolumn{2}{r}{-x^3} & +2x^2 & & \ \cline{2-3} \multicolumn{2}{r}{0} & 2x^2 & +0x & \ \multicolumn{2}{r}{} & -2x^2 & +4x & \ \cline{3-4} \multicolumn{2}{r}{} & 0 & 4x & -8 \ \multicolumn{2}{r}{} & & 4x & -8 \ \cline{4-5} \end{array} $$ Subtract $4x - 8$ from $4x - 8$ to get $0$. $$ \begin{array}{c|cc cc} \multicolumn{2}{r}{x^2} & +2x & +4 & \ \cline{2-5} x-2 & x^3 & +0x^2 & +0x & -8 \ \multicolumn{2}{r}{-x^3} & +2x^2 & & \ \cline{2-3} \multicolumn{2}{r}{0} & 2x^2 & +0x & \ \multicolumn{2}{r}{} & -2x^2 & +4x & \ \cline{3-4} \multicolumn{2}{r}{} & 0 & 4x & -8 \ \multicolumn{2}{r}{} & & -4x & +8 \ \cline{4-5} \multicolumn{2}{r}{} & & 0 & 0 \ \end{array} $$

7. Write the result

   The quotient is $x^2 + 2x + 4$, and the remainder is $0$.